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Prove that an integer of the form "$111 \dots$ upto $3^n$ digits" is divisible by $3^n$

My attempt

For $n=1,$ $111$ is divisible by 3.

Let $T_n=111...$ upto $3^n$ digits is divisible by $3^n$. Then $T_{n+1}=111...$ upto $3^{n+1}$ digits. I am unable to prove that Then $T_{n+1}$ is divisible by $3^n$.

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    $\begingroup$ Much more generally, if the base 10 digits of $m$ add up to $3^n$, then $m$ is divisible by $3^n$. e.g., the digits of 182736 add up to 27 and $182736 = 27 \times 6768$. $\endgroup$ – Robert Soupe Apr 18 '15 at 19:46
  • $\begingroup$ @RobertSoupe That is only true for n=1 and n=2. e.g. 9972 is not divisible by 27. $\endgroup$ – David P Apr 20 '15 at 5:27
  • $\begingroup$ Oops, you're right, @DavidP. Good catch. $\endgroup$ – Robert Soupe Apr 20 '15 at 12:23
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$n=1$ case: 111 is divisible by 3.

$n \Rightarrow n+1$ case: To demontrate that $T_{n+1}$ can be divided by $3^{n+1}$ I write $T_{n+1}$ as: $$ T_{n+1} = T_n \cdot (10^{2 \cdot 3^n}+10^{3^n}+1) $$ $T_n$ is divisibile by $3^n$ by induction hypotesis, $10^{2 \cdot 3^n}+10^{3^n}+1$ is divisible by 3 because the sum of digits is 3.

So $T_{n+1}$ is divisible by $3^n \cdot 3 = 3^{n+1}$ which is what we wanted.


Why $T_n$ is that one? What we want to do is splitting $T_{n+1}$ to obtain an expression of $T_n$. We can do it by splitting the number in three groups of ones. So $\underbrace{111 \dots 111}_{3^n \text{ digits}} \underbrace{111 \dots 111}_{3^n \text{ digits}} \underbrace{111 \dots 111}_{3^n \text{ digits}} = \underbrace{111 \dots 111}_{3^n \text{ digits}} \underbrace{000 \dots 000}_{3^n \text{ digits}} \underbrace{000 \dots 000}_{3^n \text{ digits}} + \underbrace{111 \dots 111}_{3^n \text{ digits}} \underbrace{000 \dots 000}_{3^n \text{ digits}} + \underbrace{111 \dots 111}_{3^n \text{ digits}}$

Formally it's done by writing the expression $T_{n+1} = T_n \cdot (10^{2 \cdot 3^n}+10^{3^n}+1)$

For example we can observe that $$T_2 = 111\,111\,111 = 111 \cdot 10^6 + 111 \cdot 10^3 + 111 = 111 \cdot (10^6+10^3+1) = \\ =T_1 \cdot (10^6+10^3+1)$$

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  • $\begingroup$ Thank you for the answer. Would you kindly elaborate in details in respect of $$ T_{n+1} = T_n \cdot (10^{6 n}+10^{3 n}+1) $$ $\endgroup$ – user1942348 Apr 18 '15 at 17:39
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    $\begingroup$ Shouldn't that be $T_n(1+10^{3^n}+10^{2\cdot 3^n})$? $\endgroup$ – zhw. Apr 18 '15 at 17:53
  • $\begingroup$ correct! i'll fix. $\endgroup$ – Blex Apr 18 '15 at 17:56
  • $\begingroup$ @Blex $$T_2 =...=T_1 \cdot (10^6+10^3+1)$$ is understandable. $$ T_{n+1} = T_n \cdot (10^{2 \cdot 3^n}+10^{3^n}+1) $$ is also coming from the same reason but can it (the generalization that $T_{n+1} = T_n \cdot (10^{2 \cdot 3^n}+10^{3^n}+1)$) be little more explainable so that beginners can catch it more easily. Thanks for your reply. $\endgroup$ – user1942348 Apr 18 '15 at 18:06
  • $\begingroup$ fixed, i hope it's better now; if it's not enought understandable, i'll write simpler. $\endgroup$ – Blex Apr 18 '15 at 18:16
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step 1. Write the number K = 111... in decimal form as 1*100 + 1*101 + 1*102 +... = $\sum_{i=0}^m10^i$

step 2. Examine K mod $3^n$ = $\sum_{i=0}^m10^i$ mod $3^n$ = $\frac{1-10^{m+1}}{1-10}$ mod $3^n$ = $\frac{10^{m+1}-1}{9}$ mod $3^n$ = $\frac{10^{3^n+1} -1}{9}$ mod $3^n$. Where the first equality follows from the formula for the sum of a finite geometric series. Clearly, 9|$10^{3^n+1} -1$ is necessary, so compute K' = $10^{3^n+1} -1$ mod (9*$3^n$)= $10^{3^n+1} -1$ mod $3^{n+2}$. If K' = 0, we're done.

step 3. $\phi$($3^{n+2}$) = $3^{n+1}$-$3^n$ = a

Thus, by Euler's formula, $10^a$ = 1 mod $3^{n+2}$. Therefore, K' + 1 = $(10^a)^{3^n}$= 1 mod $3^{n+2}$.

So, $10^{3^n +1}$ -1 mod $3^{n+2}$ = 0.

The End

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